What is the integral of e^{6t}cos(2t) ?

The integral of e^{6t}cos(2t) is (e^{6t}sin(2t))/(20)+(3e^{6t}cos(2t))/(20)+C

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integral of e^{6t}cos(2t)

\int e^{6 t} cos (2 t) d t

\frac{e ^{6 t} sin ( 2 t )}{20} + \frac{3 e ^{6 t} cos ( 2 t )}{20} + C

Solution steps

\int e^{6 t} cos (2 t) d t

Apply Integration By Parts

: \frac{1}{2} e^{6 t} sin (2 t) - \int 3 e^{6 t} sin (2 t) d t

= \frac{1}{2} e^{6 t} sin (2 t) - \int 3 e^{6 t} sin (2 t) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} e^{6 t} sin (2 t) - 3 \cdot \int e^{6 t} sin (2 t) d t

Apply Integration By Parts

: - \frac{1}{2} e^{6 t} cos (2 t) - \int - 3 e^{6 t} cos (2 t) d t

= \frac{1}{2} e^{6 t} sin (2 t) - 3 (- \frac{1}{2} e^{6 t} cos (2 t) - \int - 3 e^{6 t} cos (2 t) d t)

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} e^{6 t} sin (2 t) - 3 (- \frac{1}{2} e^{6 t} cos (2 t) - (- 3 \cdot \int e^{6 t} cos (2 t) d t))

Therefore

\int e^{6 t} cos (2 t) d t = \frac{1}{2} e^{6 t} sin (2 t) - 3 (- \frac{1}{2} e^{6 t} cos (2 t) - (- 3 \cdot \int e^{6 t} cos (2 t) d t))

Isolate

\int e^{6 t} cos (2 t) d t

= \frac{e ^{6 t} sin ( 2 t )}{20} + \frac{3 e ^{6 t} cos ( 2 t )}{20}

Add a constant to the solution

= \frac{e ^{6 t} sin ( 2 t )}{20} + \frac{3 e ^{6 t} cos ( 2 t )}{20} + C

\int_{0}^{2 π} 3 sin (2 t) d t

\frac{d}{d x} (ln (\frac{8 x}{x + 5}))

x \to 5 lim (\frac{1}{( x - 5 ) ^{3}})

\frac{\partial}{\partial x} (\frac{e ^{x}}{y + x ^{8}})

\frac{d}{d x} (\frac{x + y}{x y})

What is the integral of e^{6t}cos(2t) ?
The integral of e^{6t}cos(2t) is (e^{6t}sin(2t))/(20)+(3e^{6t}cos(2t))/(20)+C